The sequences $(x_n),(y_n),(z_n),n\in\mathbb N$, are defined by the relations
$$x_{n+1}=\frac{2x_n}{x_n^2-1},\qquad y_{n+1}=\frac{2y_n}{y_n^2-1},\qquad z_{n+1}=\frac{2z_n}{z_n^2-1},$$where $x_1=2$, $y_1=4$, and $x_1y_1z_1=x_1+y_1+z_1$.
(a) Show that $x_n^2\ne1$, $y_n^2\ne1$, $z_n^2\ne1$ for all $n$;
(b) Does there exist a $k\in\mathbb N$ for which $x_k+y_k+z_k=0$?
jasperE3 wrote:
The sequences $(x_n),(y_n),(z_n),n\in\mathbb N$, are defined by the relations
$$x_{n+1}=\frac{2x_n}{x_n^2-1},\qquad y_{n+1}=\frac{2y_n}{y_n^2-1},\qquad z_{n+1}=\frac{2z_n}{z_n^2-1},$$where $x_1=2$, $y_1=4$, and $x_1y_1z_1=x_1+y_1+z_1$.
(a) Show that $x_n^2\ne1$, $y_n^2\ne1$, $z_n^2\ne1$ for all $n$;
(b) Does there exist a $k\in\mathbb N$ for which $x_k+y_k+z_k=0$?
a) The only possible way $x_n^2=1$ was if $x_n=1$ or $x_n=-1$
Now we will Claim:- $x_n+y_n+z_n=x_n y_n z_n,\forall n$
Proof:- Just induction.
So now if $x_n^2=1,y_n^2=1,z_n^2=1$ then it would contradict our claim so it is false.
b) If $x_n+y_n+z_n=0$ then $x_n y_n z_n=0$ would imply that either $x_n$ or $y_n$ or $z_n=0$ which isn't possible so it cant equal zero.
The problem is same as a problem from the All russian math Olympiad 1990,however it is a easier version.